Z-values and probability

Suppose concentration of arsenic (in ppb) in groundwater sampled quarterly from a particular well can be modeled with a normal distribution with a mean of 5 and a standard deviation of 1.

a) What is the probability that a single observation will be given greater than 7ppb?

\(z =5-7 / 1 = 2\) And since we want greater than the probability would be 2.28.


What I'm confused on is the second part!
b)What is the probability that the average of four observations will be greater than 7 ppb?

So now, I have to take into account n, so the formula shifts from
\(z =Xi-7/ sd \) to \(z =xbar-7 / [sd/sqrt(n)] \)

\(z =5-7 /[1/sqrt(4)]= -4 \) Therefore, the probability is 100?
Which doesn't make sense at all.
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